Upper and lower bounds - Edexcel GCSE Maths (Calculator) Exam Questions

Practise 36 exam-style questions on Upper and lower bounds for Edexcel GCSE Maths (Calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

36 questions
1–5 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

A rectangle has length 12.4 cm and width 8.7 cm.
Both measurements are correct to 1 decimal place.

Work out the upper bound for the perimeter of the rectangle.

Model answer

Upper bounds: 12.45 cm and 8.75 cm
Upper bound for the perimeter = \(2 \times (12.45 + 8.75) = 2 \times 21.2 = 42.4\) cm

Explanation

To get the largest possible perimeter, use the upper bound of each length (0.05 cm more than the value given). Working out the perimeter first (42.2 cm) and then adding 0.05 is a common error - each of the four sides carries its own rounding error.

Mark scheme

1 mark
M1 for using 12.45 and 8.75
correct answer with no working scores full marks
1 mark
A1 for 42.4 cm
cao

Question 2

3 marks

A metal block has a mass of 640 g, correct to the nearest 10 g.
Its volume is 82 cm³, correct to the nearest cm³.

Work out the lower bound for the density of the metal.
Give your answer correct to 3 significant figures.

Model answer

Lower bound of mass = 640 − 5 = 635 g
Upper bound of volume = 82.5 cm³
Lower bound of density = \(635 \div 82.5 = 7.6969...\)
Lower bound = 7.70 g/cm³ (3 s.f.)

Explanation

Density = mass ÷ volume. The smallest density comes from the smallest mass divided by the largest volume. Notice the mass is to the nearest 10 g, so its bounds are 5 g either side, but the volume's bounds are only 0.5 cm³ either side.

Mark scheme

1 mark
B1 for 635 or 82.5 seen
exact bounds; correct answer with no working scores full marks
1 mark
M1 for \(635 \div 82.5\)
lower bound of mass ÷ upper bound of volume
1 mark
A1 for 7.70 g/cm³
awrt 7.70 (7.6969...); accept 7.7

Question 3

3 marks

The mass of the Earth is \(5.97 \times 10^{24}\) kg, correct to 3 significant figures.
The volume of the Earth is \(1.08 \times 10^{21}\) m³, correct to 3 significant figures.

Work out the lower bound for the density of the Earth.
Give your answer in kg/m³, correct to 3 significant figures.

Model answer

Lower bound of the mass = \(5.965 \times 10^{24}\) kg
Upper bound of the volume = \(1.085 \times 10^{21}\) m³
Lower bound of the density = \(\dfrac{5.965 \times 10^{24}}{1.085 \times 10^{21}} = 5497.69...\)
Lower bound = 5500 kg/m³ (3 s.f.), or \(5.50 \times 10^{3}\) kg/m³

Explanation

For numbers in standard form, the bounds are half a unit of the last significant figure either side: 5.97 has bounds 5.965 and 5.975 (times \(10^{24}\)). The smallest density uses the smallest mass and the largest volume. Use the EXP or ×10ˣ key so the powers of 10 are handled correctly.

Mark scheme

1 mark
B1 for \(5.965 \times 10^{24}\) or \(1.085 \times 10^{21}\) seen
exact bounds; correct answer with no working scores full marks
1 mark
M1 for \(\dfrac{5.965 \times 10^{24}}{1.085 \times 10^{21}}\)
lower bound of mass ÷ upper bound of volume
1 mark
A1 for 5500 kg/m³
awrt 5500 (5497.69...); accept \(5.50 \times 10^{3}\)

Question 4

4 marks

A cylindrical tin has radius 4.2 cm, correct to 1 decimal place.
The height of the tin is 11 cm, correct to the nearest centimetre.

The label on the tin says it holds at least 600 ml.

Can you be certain that the label is correct?
You must show how you get your answer.

Model answer

Lower bounds: radius 4.15 cm, height 10.5 cm
Lower bound of the volume = \(\pi \times 4.15^2 \times 10.5 = 568.11...\) cm³ = 568.1 ml
568.1 ml < 600 ml, so the tin might hold less than 600 ml.
We cannot be certain that the label is correct.

Explanation

To be certain the tin holds at least 600 ml, even the smallest possible volume must be at least 600 ml. Using the given values gives about 610 ml, which looks fine, but the lower bound is only about 568 ml. 1 cm³ is the same as 1 ml.

Mark scheme

1 mark
B1 for 4.15 and 10.5 seen
exact bounds
1 mark
M1 for \(\pi \times 4.15^2 \times 10.5\)
lower bound of the volume
1 mark
A1 for 568.1 cm³
awrt 568 (568.11...)
1 mark
C1 for No, because the lower bound of the volume, 568 ml, is less than 600 ml
oe; 1 cm³ = 1 ml

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